Illustrated Physics
physics, drawn
Concept

Static friction — where it appears

The force a stationary contact supplies to prevent sliding, an inequality that gives exactly what equilibrium asks until it cannot. It has no value of its own until the moment of slipping, so the coefficient describes the corner of a response curve rather than the whole of it.

Named by 7 essays across one field — each of them below, with the objects they name alongside it.

What friction returns, against what it is asked for. The friction force on a block under a 50 N normal load, against the force applied to it. Below 30.0 N — the static limit μs·N — friction returns exactly what is asked for and nothing moves, so the curve is the 45° line and the coefficient never appears. At that point the surface gives way and the force drops to μk·N = 22.5 N, where it stays however hard the block is pushed. The gap above the flat line is the surplus that accelerates it: 22.5 N at the right-hand edge of the axis.

The force that takes what it needs

Static friction has no value of its own. It supplies exactly what equilibrium demands and not a newton more, right up to the moment it cannot — which is the only instant in the whole business at which a coefficient of friction means anything at all.

mechanics · Friction Which failure comes first. The dividing line between a block that slides and one that tips over, for a horizontal push applied at the top. Sliding needs a force of μ_s times the weight; tipping needs the push's moment about the leading bottom edge to beat the weight's, which is the weight times half the width. The weight appears in both and cancels, so the boundary is the curve aspect ratio = 1/2μ_s and nothing else: not the mass, not how hard the block is pushed, and not what it is made of except through μ. At μ_s = 0.5 the dividing shape is as tall as it is wide; at μ_s = 0.2 it is 2.5 times as tall. Of the 5 objects marked, 3 sit above the line and go over rather than sliding: a paperback, standing, a full filing cabinet, a pint glass.

Slide or topple

Push a wardrobe and it goes over; push a brick and it skids. Both are held by the same friction and both are pushed by the same hand, and which of the two failures arrives first has nothing to do with how hard the push is. The floor decides it, by shifting where it pushes back.

mechanics · Free-body How much a wrap holds, against how many turns it is. The ratio of the two tensions a rope can hold across, against the number of turns it is wrapped, for coefficients of 0.1, 0.25, 0.5. The axis is logarithmic because the law is exponential, so each line is straight and its slope is the coefficient. At µ = 0.1 one turn multiplies by 1.9, two turns by 4 and three by 7 — so a person pulling with the strength of one arm holds a load that a small crane would be needed to lift. The practical consequence is the one a sailor states as a rule: turns are cheap and each is worth as much as the one before it, which is a statement about a constant factor rather than a constant force.

The part of the wrap that is actually gripping

The capstan equation gives the largest tension ratio a wrap can hold, and almost nothing spends its life at that limit. Below it the wrap divides in two: an idle arc doing nothing at all and an active arc creeping and carrying the whole exponential — and the division explains a belt's speed loss and its squeal.

mechanics · Friction The shear a pushed contact carries, and where it gives. Across a Hertzian contact of radius a, the normal pressure as a fraction of its peak, which is also the friction limit μp in units of μp₀ (dashed), and the shear traction for a sideways push of 0.3, 0.6, 0.9 times the load that would slide it, μP. Where the shear reaches the dashed ceiling the surfaces slip; inside, they stick. At 0.3 μP the stuck disc has a radius of 0.89a, so 21 per cent of the contact area is slipping; at 0.6 μP the stuck disc has a radius of 0.74a, so 46 per cent of the contact area is slipping; at 0.9 μP the stuck disc has a radius of 0.46a, so 78 per cent of the contact area is slipping. The shear is highest just inside the slipping ring and, in an elastic contact that simply stuck everywhere, would be infinite at the rim: slip at the edge is what removes that singularity.

The contact that slips before it slides

Static friction is supposed to hold perfectly until the load reaches μ times the normal force and then give way. Two elastic bodies pressed together do something else. Push a ball sideways on a flat with any force at all and a ring at the edge of their contact is already slipping, while the centre stays stuck. As the push grows the ring spreads inward, and gross sliding is only the moment the stuck centre shrinks to nothing. Shake the push back and forth and the ring slips each way on every cycle. The contact then loses energy though it never slides, as the cube of how hard it is shaken.

mechanics · Friction Two ways for a swing to die. A mass on a spring released 10.6 dead-band widths from rest, against time in units of 1/ω. Red: sliding on a dry surface, whose friction is a constant force; each swing is exactly a cosine, and the turning points fall in a straight line by two dead-band widths per half-cycle until the mass stops, 2.5 cycles after release, at −0.60. Blue: the same mass with viscous damping (damping ratio 0.05), whose force is proportional to speed; the swings shrink by the same fraction each cycle, along an exponential, and never stop. The period is the same in both and the same as with no damping at all to within the viscous case's 0.13 per cent.

The swing that dies in a straight line

A pendulum in air loses the same fraction of its swing every cycle, so its amplitude falls along an exponential and, in principle, never quite reaches zero. A mass on a spring sliding over a dry surface does something else. Its friction is a fixed force, so it loses a fixed amount of amplitude every half-swing — the swings shrink in a straight line — and it stops completely, after a countable number of swings, somewhere inside a band around its natural position whose width is the friction divided by the stiffness. Where in that band it stops depends on where it started, which is why a barometer with friction in it is tapped before it is read.

mechanics · Friction Where the thread's line meets the floor. A spool resting on the floor on its rims of radius R, its thread wound round a hub of radius r = 0.5R and leaving the hub's underside tangentially, pulled up and to the right at 20°, 60° and 78° above the horizontal; the dashed lines continue each thread's line back down to the floor. The spool turns about the contact point, and the sense of the turn is set by where the thread's line meets the floor. At 20° it meets the floor behind the contact point, on the far side from the hand, and the spool rolls towards the hand. At 60°, where cos θ = r/R, it passes exactly through the contact point; the pull has no turning effect there, and the spool stays still while the hand pulls. At 78° it meets the floor in front of the contact point, between the spool and the hand, and the spool rolls away from the hand, paying the thread out.

The spool that rolls towards the hand pulling it

Pull gently on the thread of a cotton reel lying on its side on a table, with the thread coming off the underside of the hub, and the reel rolls towards the hand, winding the thread up as it comes. Raise the thread and at one particular angle the reel will not move at all however hard it is pulled; raise it further and the reel rolls away, paying the thread out. Which way it goes is decided by a single line — where the thread's direction, extended, meets the table — and the same line decides which way a bicycle goes when its lower pedal is pulled backwards.

mechanics · Free-body A floor that has to hit a rod nobody dropped. The normal force on the lower end of a uniform rod at 63° with μ = 1.6, sliding outward at √(gl) and not yet turning, against time on a logarithmic axis, for a floor whose contact stiffness is 10⁴, 10⁵ and 10⁶ times mg/l. The rigid equations have no solution here, and the compliant floor shows why: the end digs in, the deeper it digs the harder friction twists it further down, and the force climbs until the sliding stops. The stiffer the floor, the higher and briefer the spike — 20, 56, 163 times the rod's weight, ending after 41, 18, 7 thousandths of √(l/g) — a collision with no approach.

The rod that friction cannot decide about

Newton's laws, a rigid rod and Coulomb's law of friction are the first things a mechanics course teaches, and in 1895 Paul Painlevé showed that together they can fail to give any answer at all. A rod sliding on a rough enough floor, at the right angle, has no motion consistent with the three — or has two. The way out is not a new law but a forgotten fact: real contacts give a little. With that, the rod jams, and the floor delivers a sudden blow to a rod that nothing struck — an impact without a collision, which is how a stick of chalk comes to hop.

mechanics · Friction

Named alongside it

The objects these essays reach for when they reach for this one.

FrictionContact areaCoefficient of frictionEquilibriumNormal forceDampingFree-body diagramHysteresisMechanical advantageMoment of inertiaTorqueCentre of gravity

All concepts